lim[x→0] ((a^x - 1)/x) = lim[x→0] (((e^log(a))^x - 1)/x) = lim[x→0] ((e^(x log(a)) - 1)/x) = lim[x→0] (log(a) ((e^(x log(a)) - 1)/(x log(a)))) ‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥ ‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥ あとは自力で... lim[x→0] ((a^x^2 - 1)/x) とは、 lim[x→0] (((a^x)^2 - 1)/x) か lim[x→0] ((a^(x^2) - 1)/x) のどちらの意味でしょうか? lim[x→0] (((a^x)^2 - 1)/x) = lim[x→0] ((a^(2x) - 1)/x) = lim[x→0] (2 ((a^(2x) - 1)/(2x))) ‥‥
![lim(x→0)(e^x-1)/x=1を利用して、極限値を求める問題です。①lim(x→0)(a^x-1)/x(a>0)、②lim(x→0... - Yahoo!知恵袋](https://cdn-ak-scissors.b.st-hatena.com/image/square/1d07bee2b75b182ba712690f3a3464c29972e28b/height=288;version=1;width=512/https%3A%2F%2Fs.yimg.jp%2Fimages%2Fks%2Fclap%2Fimage%2Fogp%2Fogp.png)